Hi Joseph,
Your first question relates to material that I think we will get to in
lecture on Monday, so I will bracket it for the moment. The second question
raises a number of good issues. The first is: what can I cut from and leave
from likelihoods and log-likelihoods. From our derivitation of the concept
of likelihood, all we said that we know is a function which is proportional
to the likelihood, usually phrased as:
L(theta|y) = k(y)f(y|theta)
\propto f(y|theta).
But because the actual value of the likelihood function itself has no
intrinsic meaning (we can only use likelihood ratios across proposed
parameter values) it is perfectly permissible to drop any constants which
are multiplied times the likelihood. For example, if the L(theta|y) is
proportional to a*f*y|theta) where a is some constant, we can drop 'a' and
we'll still get the same MLE. We have to be careful about applying this
rule to the log scale, however. Notice that if
L(theta|y) \propto a*f(y|theta)
then
lnL(theta|y) \propto ln(a) + ln f(y|theta)
Thus, on the log-scale we can drop additive constants and still have a valid
likelihood. But we can't drop constants which are part of the second term
e.g. if we had
ln L(theta|y) \propto ln(a) + b*ln f(y|theta)
we could drop only drop ln(a) and still retain a proportional likelihood: we
could not drop 'b'. Another way to think about this is that on the log
scale, likelihood ratios become log-likelihood differences, and dropping
multiplicative constant will clearly alter log-likelihood differences. But
additive constants won't.
You can also confirm for yourself that the second derivative won't be
affected by adding or dropping additive constants for the log-likelihood
(they just disappear when the first derivative is taken anyway) but changing
a mulitplicative constant for a log-likelihood will alter the second
derivative, and so lead to a mistaken sense of uncertainty about the MLE.
Iain
On Thu, Feb 18, 2010 at 12:09 PM, Gavinlertvatana, Poj <
pgavinlertvatana at hbs.edu> wrote:
Is the negative inverse of the hessian = the variance
of the MLE estimator?
If we remove constants in likelihood functions, take second derivatives to
get the hessian, doesn't that mean that depending on the constants we have
in the likelihood function, we get different variances on the MLE estimator?
Best regards,
Joseph
Joseph Poj Gavinlertvatana
Doctoral student, Marketing
Harvard Business School
203 Wyss Hall, Soldiers Field, Boston, MA 02163
Ph 617.230.5907
Fx 617.496.4397
Txt/Vm 617.910.0563
Em pgavinlertvatana at
hbs.edu
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